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Python Python Basics (2015) Letter Game App Even or Odd Loop

Posted by Lynn Overall
Lynn Overall
1,256 Points

while start: if numb=even_odd(random.randint(1,99)): print("{} is even").format(numb) start -=1

I know this will work so what is missing

even.py 
import random
def even_odd(num):
    # If % 2 is 0, the number is even.
    # Since 0 is falsey, we have to invert it with not.
    return not num % 2
start=5
while start:
    if numb=even_odd(random.randint(1,99)):
        print("{} is even").format(numb)
    start -=1

1 Answer

Steve Hunter
Steve Hunter
57,712 Points

Hi Lynn,

Most of that is fine; good work. But it's not quite there.

Try this ...

Import random.

Set start to 5.

Create a while loop that runs while start isn't false.

Inside the loop assign a random number to a variable. Call it numb for the hell of it.

Test if the variable, numb, is even/odd by passing it to the provided method.

If it is even print the even string, else print the odd string. Don't close the print parenthesis prior to calling .format() on the string:

print("{} insert var").format(var) # wrong
print("{} insert var".format(var)) # correct

Deduct 1 from start.

That should see you right - else shout.

Let me know how you get on.

Steve.